为什么是（A | B），相当于一个 - （一和b）+ B？ [英] Why is (a | b ) equivalent to a - (a & b) + b?

问题描述

我一直在寻找一种方式做一个BITOR（）与Oracle数据库并在一个建议来只使用BITAND（）代替，替换BITOR（A，B）与A + B - BITAND（A，B）

I was looking for a way to do a BITOR() with an Oracle database and came across a suggestion to just use BITAND() instead, replacing BITOR(a,b) with a + b - BITAND(a,b).

我手工测试了几次，并验证它似乎所有的二进制数我能想到的工作，但我想不出为什么这是正确的了快速的数学证明。

有人能告诉我吗？

I tested it by hand a few times and verified it seems to work for all binary numbers I could think of, but I can't think out quick mathematical proof of why this is correct.
Could somebody enlighten me?

推荐答案

A和B是位​​集是在A和B. A的 - （A和B），让你与所有那些只在A.添加B到了位，你会得到一切都在位在A或那些在B.

A & B is the set of bits that are on in both A and B. A - (A & B) leaves you with all those bits that are only on in A. Add B to that, and you get all the bits that are on in A or those that are on in B.

简单添加A和B都不会因为携带在哪里工作都具有1位。首先取出A和B共同位，我们知道，（A-（A和B））将与B公共无位，所以加在一起保证不会产生套利

Simple addition of A and B won't work because of carrying where both have a 1 bit. By removing the bits common to A and B first, we know that (A-(A&B)) will have no bits in common with B, so adding them together is guaranteed not to produce a carry.

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